Evaluate $\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt$ . $$\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt$$
To determine the above value, do I need to integrate first? Or, is there some other way to do this?
 A: Let $f(x)=\int_0^x\dfrac{t^2}{t^4+1}dt$
Now you have to compute $\lim\limits_{x\to 0}\dfrac{f(x)}{x^3}$.
This can be done in a hospital !
A: Alternatively, expand the integrand to Taylor series:
$$\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt=$$
$$\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x t^2(1-t^4+t^8-...)dt=$$
$$\lim\limits_{x\to 0}\frac{\frac{x^3}{3}-\frac{x^7}{7}+O(x^{11})}{x^3}=\frac13.$$
A: This is an application of L'Hôpital's Rule because 
$$ \lim_{x\to 0}\int_0^x\frac{t^2}{t^4+1}dt=0$$ 
and
$$ \lim_{x\to 0}x^3=0.$$
So, we have:
$$ \lim_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt=\lim_{x\to 0}\frac{\int_0^x \frac{t^2}{t^4+1}dt}{x^3}\stackrel{\text{LH}}{=}\lim_{x\to 0}\frac{x^2}{3x^2(x^4+1)}=\lim_{x\to 0}\frac{1}{3(x^4+1)}=\frac{1}{3}.$$
A: The integrand $f(t)=\frac{t^2}{1+t^4}$ satisfies the inequality $(1-\epsilon)(t^2)\le f(t)\le t^2$ for $0\le t \le x$ when $x$ is sufficiently small and positive, with $\epsilon$ any small positive number.  Therefore by comparison we find that the full expression in the problem lies between $\frac{1-\epsilon}{3}$ and $1/3$ forcing the limit to be $1/3$ as $x$ approaches zero from above.  A similar comparison also gives the limit $1/3$ as $x$ approaches zero from below making the limit well defined and equal to $1/3$.
A: Note that the function under limit is an even function and hence it is sufficient to consider the limit as $x\to 0^{+}$. So in what follows we assume that $x$ is positive.
Note that if $0\leq t\leq x$ then we have $$\frac{t^{2}}{x^{4}+1}\leq\frac{t^{2}}{t^{4}+1}\leq t^{2}$$ and integrating the above with respect to $t$ in interval $[0,x]$ and dividing the resulting inequality by $x^{3}$ we get $$\frac{1}{3(x^{4}+1)}\leq\frac{1}{x^{3}}\int_{0}^{x}\frac{t^{2}}{t^{4}+1}\,dt\leq\frac{1}{3}$$ and now applying Squeeze theorem we get the desired limit as $1/3$.
